A curve in the plane is defined parametrically by the equations $x=7\ln(t)$ and $y=\sqrt{1-4t}$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{\sqrt{1-4t}}{14t}$ (Choice B) B $-\dfrac{2t}{7\sqrt{1-4t}}$ (Choice C) C $-\dfrac{2}{\sqrt{1-4t}}$ (Choice D) D $\dfrac{t}{14\sqrt{1-4t}}$
In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=7\ln(t)$ and $y=\sqrt{1-4t}$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}\left(\sqrt{1-4t}\right)}{\dfrac{d}{dt}(7\ln(t))} \\\\ &=\dfrac{\dfrac{-2}{\sqrt{1-4t}}}{\dfrac{7}{t}} \\\\ &=-\dfrac{2t}{7\sqrt{1-4t}} \gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=-\dfrac{2t}{7\sqrt{1-4t}}$.